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Expressive Words
Sometimes people repeat letters to represent extra feeling. For example:

- "hello" -> "heeellooo"
- "hi" -> "hiiii"
In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.

- For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has a size less than three. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If s = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = s.
Return the number of query strings that are stretchy.

Example 1:

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Example 2:

Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3

Constraints:

- 1 <= s.length, words.length <= 100
- 1 <= words[i].length <= 100
- s and words[i] consist of lowercase letters. 

/**
 * @param {string} s
 * @param {string[]} words
 * @return {number}
 */
func expressiveWords(s string, words []string) int {
 var convertStrToCharMap = func(str string) map[string][]int {
  lastChar := ""
  charMap := make(map[string][]int)

  for _, char := range strings.Split(str, "") {
   if char != lastChar {
    charMap[char] = append(charMap[char], 1)
   } else {
    charMap[char][len(charMap[char])-1]++
   }
   lastChar = char
  }
  return charMap
 }

 inputCharMap := convertStrToCharMap(s)

 checkWord := func(charMap map[string][]int, word string) bool {
  wordCharMap := convertStrToCharMap(word)
  for key := range charMap {
   if len(charMap[key]) != len(wordCharMap[key]) {
    return false
   }

   for i := 0; i < len(charMap[key]); i++ {
    if charMap[key][i] < wordCharMap[key][i] ||
     (charMap[key][i] < 3 && charMap[key][i] != wordCharMap[key][i]) ||
     wordCharMap[key][i] == 0 {
     return false
    }
   }
  }
  return true
 }

 output := 0
 for _, word := range words {
  var wg sync.WaitGroup
  wg.Add(1)
  go func(w string) {
   defer wg.Done()
   if checkWord(inputCharMap, w) {
    output++
   }
  }(word)
  wg.Wait()
 }
 return output
}

呢題唔識做 寫得好差

Running Sum of 1d Array

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]

Constraints:

1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

Answer:

/**
 * @param {number[]} nums
 * @return {number[]}
 */
const runningSum = (nums)=>{
  const numsLength = nums.length
  return nums.reduce((result, curr, index)=>{
     result.push(curr+result[index])
     return result
  }, [0]).slice(1)  
};

JS reduce 很好用

/**
 * @param {number[]} nums
 * @return {number[]}
 */
const runningSum = (nums)=>{
  return nums.reduce((result, curr, index)=>{
     result.push(curr+result[index])
     return result
  }, [0]).slice(1)  
};

有行del少咗

Reduce個initial value 呀嘛
用咗result第一次行會係[0]開始
如果唔係[0]第一次行result會係由input第一個value開始
input係[int,int...] 所以result會係一個int 而int[index]會係undefine
我又唔想reduce行緊嗰時每次check result[index]係唔係undefine 所以俾個0佢之後slice返

const runningSum = (nums)=>{
  let result = []
  nums.reduce((prev, curr)=>{
     result.push(prev+curr)
  }, 0)
  return result
};

其實現實咁寫會好啲 我貪快
不過最後睇人地做用map原來一行做到 仲簡單

const runningSum = (nums)=>{
  let result = []
  nums.reduce((prev, curr)=>{
     const sum=prev+curr
     result.push(sum)
     return sum
  }, 0)
  return result
};

咁先啱

forloop快好多 啱啱試咗呢題快20%左右

Transpose Matrix
Given a 2D integer array matrix, return the transpose of matrix.

The transpose of a matrix is the matrix flipped over its main diagonal, switching the matrix's row and column indices.

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[1,4,7],[2,5,8],[3,6,9]]

Example 2:

Input: matrix = [[1,2,3],[4,5,6]]
Output: [[1,4],[2,5],[3,6]]

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 1000
1 <= m * n <= 105
-109 <= matrix[i][j] <= 109

Answer:

func transpose(matrix [][]int) [][]int {
 transposeOfMatrix := make([][]int, len(matrix[0]))
 for i := 0; i < len(matrix[0]); i++ {
  for j := 0; j < len(matrix); j++ {
            transposeOfMatrix[i] = append(transposeOfMatrix[i], matrix[j][i])
  }
 }
 return transposeOfMatrix
}

順手做埋